3070 Fibonacci 矩阵乘法

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3502		Accepted: 2450

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

Stanford Local 2006

 

 

#include
int n;
int a[4],b[4];
const int M=10000;
void cal(int n)
{
    if(n==1||n==0)
    {
        a[3]=1;a[2]=1;a[1]=1;a[0]=0;
        return ;
    }
    if(n&1)
    {
        cal(n/2);
        b[3]=(a[3]*a[3]+a[2]*a[1])%M;
        b[2]=(a[3]*a[2]+a[2]*a[0])%M;
        b[1]=(a[1]*a[3]+a[0]*a[1])%M;
        b[0]=(a[1]*a[2]+a[0]*a[0])%M;
        for(int i=0;i<4;i++)  a[i]=b[i];
        b[3]=(a[3]+a[2])%M;
        b[2]=a[3];
        b[1]=(a[1]+a[0])%M;
        b[0]=a[1];
        for(int i=0;i<4;i++)  a[i]=b[i];
    }
    else
    {
        cal(n/2);
        b[3]=(a[3]*a[3]+a[2]*a[1])%M;
        b[2]=(a[3]*a[2]+a[2]*a[0])%M;
        b[1]=(a[1]*a[3]+a[0]*a[1])%M;
        b[0]=(a[1]*a[2]+a[0]*a[0])%M;
        for(int i=0;i<4;i++)  a[i]=b[i];
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)  break;
        if(n==0) {printf("0/n");continue;}
        cal(n-1);
        printf("%d/n",a[3]);
    }
    return 0;
}